E作者kcuricky (天鹅)
看板trans_math
标题Re: [单变]微积分
时间Fri Sep 8 23:39:07 2006
※ 引述《GBRS ()》之铭言：
: Let f(x) be a differentialbe function on |R satisfying
: ╭x^2
: f(x^2)=1+│ f(y)( 1-tan y )dy for all xε|R
: ╯0
: Then f(π)=?
f'(x^2)*2x=f(x^2)(1-tanx^2)*2x
f'(x^2)=f(x^2)(1-tanx^2)
设x^2=k
f'(k)=f(k)(1-tank)
f'(k)-(1-tank)f(k)=0
设g(k)=[e^-(k+1/2lncos^2k)]f(k)
g'(k)=-[e^(k+1/2lncos^2k)]*(1-tank)f(k)+[e^(k+1/2lncos^2k)]f'(k)
g'(k)=e^(k+1/2lncos^2k)(f'(k)-(1-tank)f(k))=0
即g(k)为常数=[e^-(k+1/2lncos^2k)]f(k)=u
又由题设知f(0)=1=k*e^0
k=1
f(k)=e^(k+1/2lncos^2k)
f(π)=e^(π+1/2lncos^2π)
=e^π
ans:e^π
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◆ From: 220.130.182.146
※ 编辑: kcuricky 来自: 220.130.182.146 (09/08 23:40)